Three-manifold surgery presentations and Kirby moves — Examples and Counterexamples

Applying [A1] to the range of any sequence $(X_n)$ in [A2] and composing the resulting choice function with $n\mapsto X_n$ gives the countable-choice function required by [A2]. Thus the chosen tubular and slope interfaces used below are available.

For the empty link there are no tubular neighbourhoods to remove and no solid tori to attach. The definition [F1] therefore leaves the standard oriented $S^3$ unchanged.

Use $N(U)=\{\mathrm{\mid }{z}_2\mathrm{\mid }\le 1\mathrm{/}\sqrt{2}\}$; its exterior is the solid torus $E_U=\{\mathrm{\mid }{z}_2\mathrm{\mid }\ge 1\mathrm{/}\sqrt{2}\}\cong D^2\times S^1$, with disk coordinate $z_1$ and core coordinate $\arg z_2$. The loop in the $z_1$ direction has degree one and bounds a disk in $E_U$, so [F4]–[F5] identify it with ${\lambda }_U$; the positive meridian ${\mu }_U$ runs once around the exterior core. For $0$-surgery, [F2] sends the attached meridian to ${\lambda }_U$, the meridian of $E_U$. The gluing-map convention in the slope definition permits the product gluing: the two disk factors glue to $S^2$ and the core circles glue to $S^1$, giving $S^1\times S^2$ with the induced surgery orientation.

For $p>0$, [F2] identifies the $p$-framed unknot filling with the filling of $E_U$ along $p[{\mu }_U]+[{\lambda }_U]$. This is exactly $L(p,1)$ by [F3]. The case $p=1$ is included.

By [F1], surgery on $D_0$ leaves the standard oriented $S^3$ unchanged.

Give $S^3$ its unit-sphere model in ${\mathbb{C}}^2$ and let $U(t)=(e^{2\pi it},0)$. With $N(U)=\{\mathrm{\mid }{z}_2\mathrm{\mid }\le 1\mathrm{/}\sqrt{2}\}$, the exterior is $D^2\times S^1$ with disk coordinate $z_1$ and core coordinate $w=z_2\mathrm{/}\mathrm{\mid }{z}_2\mathrm{\mid }$. The preferred longitude ${\lambda }_U$ is the disk-meridian class and ${\mu }_U$ is the core-longitude class by [F2]–[F3]. The diffeomorphism $g(z,w)=(w^{-1}z,w)$ preserves orientation and sends $[{\mu }_U]+[{\lambda }_U]$ to $[{\mu }_U]$. Let $f_0$ be the original $S^3$ gluing, which sends the meridian of the removed unknot neighbourhood to $[{\mu }_U]$, and choose the $+1$-filling map $f_1=g^{-1}\circ f_0$. It sends the attached meridian to $[{\mu }_U]+[{\lambda }_U]$, the required slope by [F1]. The map equal to $g$ on the exterior and the identity on the attached solid torus is well-defined because $g\circ f_1=f_0$; it is a smooth orientation-preserving diffeomorphism to the original $S^3$. Thus $D_1$ also presents $S^3$.

The diagrams are not ambient isotopic: $D_0$ has zero components and $D_1$ has one, while [F5] implies ambient isotopy preserves component count. Yet they have the same oriented surgery result by steps 1.1–1.2. Moreover [F4] identifies their difference as a permitted stabilization. Hence the false claim fails.